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%I #25 Mar 26 2024 14:45:22
%S 0,5,28,123,506,2041,8184,32759,131062,524277,2097140,8388595,
%T 33554418,134217713,536870896,2147483631,8589934574,34359738349,
%U 137438953452,549755813867,2199023255530,8796093022185,35184372088808
%N a(n) = (-1/2)*Sum_{i1 + i2 + i3 = 2*n} ((2*n)!/(i1! i2! i3!))*B(i1), where B are the Bernoulli numbers (with i1, i2, i3 >= 1).
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BernoulliNumber.html">Bernoulli number</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bernoulli_number">Bernoulli number</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).
%F a(n) = 2^(2*n - 1) - n - 1.
%F From _Colin Barker_, Dec 01 2012: (Start)
%F a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3).
%F G.f.: x^2*(2*x - 5)/((x - 1)^2*(4*x - 1)). (End)
%e From _Petros Hadjicostas_, Jul 25 2020: (Start)
%e For n = 1, a(1) = 0 because there are no integers i1, i2, i3 >= 1 s.t. i1 + i2 + i3 = 2.
%e For n = 2, a(2) = 5 because we have 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 = 4, multinomial(4, [1, 1, 2]) = multinomial(4, [1, 2, 1]) = multinomial(4, [2, 1, 1]) = 12, B(1) = -1/2, and B(2) = 1/6. Thus, a(2) = (-1/2)*(12*(-1/2) + 12*(-1/2) + 12*(1/6)) = 5. (End)
%t LinearRecurrence[{6,-9,4},{0,5,28},30] (* _Harvey P. Dale_, Dec 02 2014 *)
%o (PARI) a(n)=2^(2*n-1)-n-1
%o /* Second PARI program using the definition */
%o lista(nn) = {my(a=vector(nn)); for(n=1, nn, a[n] = (-1/2)*sum(i1=1, 2*n-2, sum(i2=1, 2*n-i1-1, binomial(2*n, i1)*binomial(2*n-i1, i2)*bernfrac(i1)))); a;} \\ _Petros Hadjicostas_, Jul 25 2020
%Y Cf. A001469.
%K nonn,easy
%O 1,2
%A _Benoit Cloitre_, Nov 30 2006
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